Expansions — Test Yourself

If x² + $\dfrac{1}{\text{x}^2}$ = 3, the value of x - $\dfrac{1}{\text{x}}$ is:

1. 1

2. -1

3. $\pm 1$

4. 0

Answer

Given, x² + $\dfrac{1}{\text{x}^2}$ = 3

As we know

$\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \times x \times \dfrac{1}{x}\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 3 - 2\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 1\\[1em] \Rightarrow x - \dfrac{1}{x} = \sqrt{1}\\[1em] \Rightarrow x - \dfrac{1}{x} = \pm 1$

Hence, option 3 is the correct option.

If a³ + b³ + c³ = 3abc, then a + b + c is equal to:

1. 1

2. -1

3. $\pm 1$

4. 0

Answer

Given, a³ + b³ + c³ = 3abc

By formula,

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

⇒ 3abc - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

⇒ 0 = (a + b + c)(a² + b² + c² - ab - bc - ca)

⇒ (a + b + c) = 0 or (a² + b² + c² - ab - bc - ca) = 0

Hence, option 4 is the correct option.

$\dfrac{196 \times 196 \times 196 + 204 \times 204 \times 204}{(196)^2 + (204)^2 - 196 \times 204}$ is equal to:

1. 400

2. -8

3. 8

4. none of these

Answer

Given, $\dfrac{196 \times 196 \times 196 + 204 \times 204 \times 204}{(196)^2 + (204)^2 - 196 \times 204}$

Using the formula, (a³ + b³) = (a + b)(a² - ab + b²)

$\Rightarrow \dfrac{196^3 + 204^3}{(196)^2 + (204)^2 - 196 \times 204}\\[1em] = \dfrac{\Big(196 + 204\Big)\Big((196)^2 + (204)^2 - 196 \times 204\Big)}{(196)^2 + (204)^2 - 196 \times 204}\\[1em] = (196 + 204)\\[1em] = 400.$

Hence, option 1 is the correct option.

If a = x + $\dfrac{1}{\text{x}}$ and b = x - $\dfrac{1}{\text{x}}$, then a² - b² is:

1. x² + $\dfrac{1}{\text{x}^2}$

2. x² - $\dfrac{1}{\text{x}^2}$

3. 4

4. $2\Big(x^2 + \dfrac{1}{\text{x}^2}\Big)$

Answer

Given, a = x + $\dfrac{1}{\text{x}}$ and b = x - $\dfrac{1}{\text{x}}$

$\Rightarrow a^2 - b^2 = \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2\\[1em] = x^2 + \Big(\dfrac{1}{x}\Big)^2 + 2 \times x \times \dfrac{1}{x} - \Big[x^2 + \Big(\dfrac{1}{x}\Big)^2 - 2 \times x \times \dfrac{1}{x}\Big]\\[1em] = x^2 + \dfrac{1}{x^2} + 2 - \Big[x^2 + \dfrac{1}{x^2} - 2\Big]\\[1em] = x^2 + \dfrac{1}{x^2} + 2 - x^2 - \dfrac{1}{x^2} + 2\\[1em] = 4$

Hence, option 3 is the correct option.

Statement 1: x > 0 and $x^2 + \dfrac{1}{x^2}$ = 2, then $x^2 - \dfrac{1}{x^2}$ = 0

Statement 2: $(x + \dfrac{1}{x})^2 = x^2 + \dfrac{1}{x^2} + 2$ = 2 + 2 = 4

$(x - \dfrac{1}{x})^2 = x^2 + \dfrac{1}{x^2} - 2$ = 2 - 2 = 0

and, $(x^2 - \dfrac{1}{x^2}) = (x + \dfrac{1}{x})(x - \dfrac{1}{x})$

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given, x² + $\dfrac{1}{\text{x}^2}$ = 2

As we know,

$\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \times x \times \dfrac{1}{x}\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 2 - 2\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 0\\[1em] \Rightarrow x - \dfrac{1}{x} = \sqrt{0}\\[1em] \Rightarrow x - \dfrac{1}{x} = 0\\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2 \times x \times \dfrac{1}{x}\\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2\\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 2 + 2\\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 4\\[1em] \Rightarrow x + \dfrac{1}{x} = \sqrt{4}\\[1em] \Rightarrow x + \dfrac{1}{x} = \pm 2\\[1em]$

Now, we know that $(x^2 - \dfrac{1}{x^2}) = (x + \dfrac{1}{x})(x - \dfrac{1}{x})$

= $\pm$ 2 x 0

= 0.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Assertion (A): x² - 5x - 1 = 0

⇒ x - $\dfrac{1}{x}$ = 5 is true.

Reason (R): x² - 5x- 1 = 0

⇒ x - $\dfrac{1}{x}$ = 5

But x - $\dfrac{1}{x}$ = 5 is true when x ≠ 0.

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

Given, x² - 5x - 1 = 0

⇒ x² - 1 = 5x

⇒ $\dfrac{x^2 - 1}{x}$ = 5

⇒ $\dfrac{x^2}{x} - \dfrac{1}{x}$ = 5

⇒ x - $\dfrac{1}{x}$ = 5

So, x - $\dfrac{1}{x}$ = 5 is true when x ≠ 0

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Assertion (A): $\dfrac{ab + bc + ca - a^2 - b^2 - c^2}{3abc - a^3 - b^3 - c^3}$

= $\dfrac{1}{a + b + c}$

Reason (R): $\dfrac{ab + bc + ca - a^2 - b^2 - c^2}{3abc - a^3 - b^3 - c^3}$

= $\dfrac{a^2 + b^2 + c^2 - ab - bc - ca}{a^3 + b^3 + c^3 - 3abc}$

= $\dfrac{a^2 + b^2 + c^2 - ab - bc - ca}{(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)}$

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

We know,

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

So, 3abc − a³ − b³ − c³ = −(a³ + b³+ c³ −3abc)

= −(a + b + c)(a² + b²+ c² − ab − bc − ca)

Given, $\dfrac{ab + bc + ca - a^2 - b^2 - c^2}{3abc - a^3 - b^3 - c^3}$

= $\dfrac{-(a^2 + b^2 + c^2 − ab − bc − ca )}{−(a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca)}$

= $\dfrac{(a^2 + b^2 + c^2 − ab − bc − ca )}{(a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca)}$

= $\dfrac{1}{a + b + c}$

So, assertion (A) is true.

By formula,

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

Given expression, $\dfrac{ab + bc + ca - a^2 - b^2 - c^2}{3abc - a^3 - b^3 - c^3}$

= $\dfrac{a^2 + b^2 + c^2 - ab - bc - ca}{a^3 + b^3 + c^3 - 3abc}$

= $\dfrac{a^2 + b^2 + c^2 - ab - bc - ca}{(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)}$

So, Reason (R) is true.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

Assertion (A): If x > y, x + y = 6 and x - y = 2 then x² + y² = 40

Reason (R): (x + y)² + (x - y)² = 2(x² + y²)

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

By formula,

⇒ (x + y)² = x² + y² + 2xy ....(1)

⇒ (x - y)² = x² + y² - 2xy .....(2)

Adding equation (1) and (2), we get :

⇒ (x + y)² + (x - y)² = x² + y² + 2xy + x² + y² - 2xy

⇒ (x + y)² + (x - y)² = 2(x² + y²)

So, reason (R) is true.

Given,

x + y = 6 and x - y = 2

By formula,

⇒ 2(x² + y²) = (x + y)² + (x - y)²

⇒ 2(x² + y²) = 6² + 2²

⇒ 2(x² + y²) = 36 + 4

⇒ 2(x² + y²) = 40

⇒ x² + y² = $\dfrac{40}{2}$

⇒ x² + y² = 20.

So, assertion (A) is false.

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Assertion (A): 5³ - 3³ - 2³ = 3 x 5 x -3 x -2

Reason (R): ∵ 5 - 3 - 2 = 0

⇒ 5³ - 3³ - 2³ = 5³ + (-3)³ + (-2)³

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

By property,

If a + b + c = 0, then

⇒ a³ + b³ + c³ = 3abc

Given,

⇒ 5³ - 3³ - 2³

⇒ 5³ + (-3)³ + (-2)³

Let a = 5, b = -3 and c = -2.

∴ a + b + c = 5 + (-3) + (-2) = 5 - 3 - 2 = 0

⇒ a³ + b³ + c³ = 3abc

⇒ 5³ + (-3)³ + (-2)³ = 3 × 5 × (-3) × (-2)

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

Simplify (x + 6)(x + 4)(x - 2)

Answer

Given,

⇒ (x + 6)(x + 4)(x - 2)

⇒ (x + 6)(x² - 2x + 4x - 8)

⇒ (x + 6)(x² + 2x - 8)

⇒ x³ + 2x² - 8x + 6x² + 12x - 48

⇒ x³ + 8x² + 4x - 48.

Hence, (x + 6)(x + 4)(x - 2) = x³ + 8x² + 4x - 48.

Simplify (x - 6)(x - 4)(x + 2)

Answer

Given,

⇒ (x - 6)(x - 4)(x + 2)

⇒ (x - 6)(x² + 2x - 4x - 8)

⇒ (x - 6)(x² - 2x - 8)

⇒ x³ - 2x² - 8x - 6x² + 12x + 48

⇒ x³ - 8x² + 4x + 48.

Hence, (x - 6)(x - 4)(x + 2) = x³ - 8x² + 4x + 48.

Simplify (x - 6)(x - 4)(x - 2)

Answer

Given,

⇒ (x - 6)(x - 4)(x - 2)

⇒ (x - 6)(x² - 2x - 4x + 8)

⇒ (x - 6)(x² - 6x + 8)

⇒ x³ - 6x² + 8x - 6x² + 36x - 48

⇒ x³ - 12x² + 44x - 48.

Hence, (x - 6)(x - 4)(x - 2) = x³ - 12x² + 44x - 48.

Simplify (x + 6)(x - 4)(x - 2)

Answer

Given,

⇒ (x + 6)(x - 4)(x - 2)

⇒ (x + 6)(x² - 2x - 4x + 8)

⇒ (x + 6)(x² - 6x + 8)

⇒ x³ - 6x² + 8x + 6x² - 36x + 48

⇒ x³ - 28x + 48.

Hence, (x + 6)(x - 4)(x - 2) = x³ - 28x + 48.

Simplify using following identity;

(a $\pm$ b)(a² $\mp$ ab + b²) = a³ $\pm$ b³

(i) (2x + 3y)(4x² - 6xy + 9y²)

(ii) $\Big(\dfrac{a}{3} - 3b\Big)\Big(\dfrac{a^2}{9} + ab + 9b^2\Big)$

Answer

(i) Given,

⇒ (2x + 3y)(4x² - 6xy + 9y²)

⇒ (2x + 3y)[(2x)² - 2x × 3y + (3y)²]

Comparing above equation with (a $\pm$ b)(a² $\pm$ ab $\pm$ b²) = a³ $\pm$ b³, we get :

a = 2x and b = 3y

∴ (2x + 3y)[(2x)² - 2x × 3y + (3y)²] = (2x)³ + (3y)³

= 8x³ + 27y³.

Hence, (2x + 3y)(4x² - 6xy + 9y²) = 8x³ + 27y³.

(ii) Given,

$\Rightarrow \Big(\dfrac{a}{3} - 3b\Big)\Big(\dfrac{a^2}{9} + ab + 9b^2\Big) \\[1em] \Rightarrow \Big(\dfrac{a}{3} - 3b\Big)\Big[\Big(\dfrac{a}{3}\Big)^2 + \dfrac{a}{3} \times 3b + (3b)^2\Big]\\[1em]$

Using identity given in question :

$\therefore \Big(\dfrac{a}{3} - 3b\Big)\Big[\Big(\dfrac{a}{3}\Big)^2 + \dfrac{a}{3} \times 3b + (3b)^2\Big] = \Big(\dfrac{a}{3}\Big)^3 - (3b)^3\\[1em] = \dfrac{a^3}{27} - 27b^3.$

Hence, $\Big(\dfrac{a}{3} - 3b\Big)\Big(\dfrac{a^2}{9} + ab + 9b^2\Big) = \dfrac{a^3}{27} - 27b^3$.

Using suitable identity, evaluate :

(i) (104)³

(ii) (97)³

Answer

(i) Given,

⇒ (104)³

⇒ (100 + 4)³

Using identity :

(a + b)³ = a³ + b³ + 3ab(a + b)

⇒ (100 + 4)³ = (100)³ + 4³ + 3 × 100 × 4 × (100 + 4)

⇒ (100 + 4)³ = 1000000 + 64 + 124800

⇒ 1124864.

Hence, (104)³ = 1124864.

(ii) Given,

⇒ (97)³

⇒ (100 - 3)³

⇒ (100)³ - 3³ - 3 × 100 × 3 × (100 - 3)

⇒ 1000000 - 27 - 900 × 97

⇒ 1000000 - 27 - 87300

⇒ 912673.

Hence, (97)³ = 912673.

Simplify :

$\dfrac{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}{(x - y)^3 + (y- z)^3 + (z - x)^3}$

Answer

If a + b + c = 0; we have :

a³ + b³ + c³ = 3abc.

Since, (x² - y²) + (y² - z²) + (z² - x²) = 0.

∴ $(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3 = 3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2)$ .........(1)

Also,

(x - y) + (y - z) + (z - x) = 0

∴ $(x - y)^3 + (y- z)^3 + (z - x)^3 = 3(x - y)(y - z)(z - x)$ ..............(2)

Dividing equation (1) by (2), we get :

$\Rightarrow \dfrac{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}{(x - y)^3 + (y- z)^3 + (z - x)^3} = \dfrac{3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2)}{3(x - y)(y - z)(z - x)} \\[1em] = \dfrac{3(x - y)(x + y)(y - z)(y + z)(z - x)(z + x)}{3(x - y)(y - z)(z - x)} \\[1em] = (x + y)(y + z)(z + x).$

Hence, $\dfrac{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}{(x - y)^3 + (y- z)^3 + (z - x)^3}$ = (x + y)(y + z)(z + x).

Evaluate:

$\dfrac{0.8 \times 0.8 \times 0.8 + 0.5 \times 0.5 \times 0.5}{0.8 \times 0.8 - 0.8 \times 0.5 + 0.5 \times 0.5}$

Answer

Given,

$\Rightarrow \dfrac{0.8 \times 0.8 \times 0.8 + 0.5 \times 0.5 \times 0.5}{0.8 \times 0.8 - 0.8 \times 0.5 + 0.5 \times 0.5} \\[1em] \Rightarrow \dfrac{(0.8)^3 + (0.5)^3}{(0.8)^2 + (0.5)^2 - 0.8 \times 0.5}$

Using the formula, (a³ + b³) = (a + b)(a² - ab + b²)

$= \dfrac{(0.8 + 0.5)[(0.8)^2 + (0.5)^2 - 0.8 \times 0.5]}{(0.8)^2 + (0.5)^2 - 0.8 \times 0.5}\\[1em] = (0.8 + 0.5)\\[1em] = 1.3$

Hence, $\dfrac{0.8 \times 0.8 \times 0.8 + 0.5 \times 0.5 \times 0.5}{0.8 \times 0.8 - 0.8 \times 0.5 + 0.5 \times 0.5}$ = 1.3.

If a - 2b + 3c = 0; state the value of a³ - 8b³ + 27c³.

Answer

By property,

If x + y + z = 0, then :

x³ + y³ + z³ = 3xyz.

Given,

⇒ a - 2b + 3c = 0

⇒ a + (-2b) + 3c = 0

∴ a³ + (-2b)³ + (3c)³ = 3 × a × (-2b) × 3c

⇒ a³ - 8b³ + 27c³ = -18abc.

Hence, a³ - 8b³ + 27c³ = -18abc.

If x + 5y = 10; find the value of x³ + 125y³ + 150xy - 1000.

Answer

Given,

⇒ x + 5y = 10

Cubing both sides we get :

⇒ (x + 5y)³ = 10³

⇒ x³ + (5y)³ + 3 × x × 5y × (x + 5y) = 1000

⇒ x³ + 125y³ + 15xy × 10 = 1000

⇒ x³ + 125y³ + 150xy - 1000 = 0.

Hence, x³ + 125y³ + 150xy - 1000 = 0.

If a + b = 11 and a² + b² = 65; find a³ + b³.

Answer

By formula,

⇒ (a + b)² = a² + b² + 2ab

⇒ 11² = 65 + 2ab

⇒ 121 = 65 + 2ab

⇒ 2ab = 56

⇒ ab = $\dfrac{56}{2}$ = 28.

By formula,

⇒ (a + b)³ = a³ + b³ + 3ab(a + b)

⇒ 11³ = a³ + b³ + 3 × 28 × 11

⇒ 1331 = a³ + b³ + 924

⇒ a³ + b³ = 1331 - 924 = 407.

Hence, a³ + b³ = 407.

If x, y and z are three different numbers, then prove that :

x² + y² + z² - xy - yz - zx is always positive.

Answer

Given,

⇒ x² + y² + z² - xy - yz - zx

Multiplying the above equation by 2,

⇒ 2(x² + y² + z² - xy - yz - zx )

⇒ 2x² + 2y² + 2z² - 2xy - 2yz - 2zx

⇒ x² + x² + y² + y² + z² + z² - 2xy - 2yz - 2zx

⇒ x² + y² - 2xy + y² + z² - 2yz + z² + x² - 2zx

⇒ (x - y)² + (y - z)² + (z - x)²

From above equation we can see that for distinct value of x, y and z given equation is always positive.

Given x = $\sqrt[3]{65}$ and y = $\sqrt[3]{64}$, find the value of

$(x - y) - \dfrac{1}{x^2 + xy + y^2}$

Answer

Given,

x = $\sqrt[3]{65}$

y = $\sqrt[3]{64}$

By formula,

x³ - y³ = (x - y) (x² + xy + y²)

x³ - y³ = $(\sqrt[3]{65})^3$ - $(\sqrt[3]{64})^3$ = 65 - 64 = 1

∴ 1 = (x - y) (x² + xy + y²)......(1)

Solving,

$\Rightarrow (x - y) - \dfrac{1}{x^2 + xy + y^2} \\[1em] \Rightarrow \dfrac{(x^2 + xy + y^2)(x - y) - 1}{x^2 + xy + y^2} \\[1em] \Rightarrow \dfrac{1 - 1}{x^2 + xy + y^2} \text{ ..[From equation (1)]} \\[1em] \Rightarrow 0.$

Hence, $(x - y) - \dfrac{1}{x^2 + xy + y^2}$ = 0.